# Solution to diff. eqn as a ring

1. Sep 29, 2008

### Alphaboy2001

Solution to diff. eqn as a ring(killing me - please look at my result!)

1. The problem statement, all variables and given/known data

I am presented with the following problem given the direction field

$$f(u,v) = \left( \begin{array}{ccc} -sin(u) & sin(v) \\ cos(u) & cos(v) \end{array} \right)$$

as $$\left( \begin{array}{c}u^{'} & v^{'} \end{array} \right) = f(u,v)$$

show that

$$(\{(u,v) \in \mathbb{R}^2|f(u,v) = 0\} = T_{1} \cup T_{2} )$$

where

$$T_{1} = \{l \pi, m\pi + \frac{\pi}{2}|l,m \in \mathbb{Z}\}$$
$$T_{2} = \{m \pi, l\pi + \frac{\pi}{2}|l,m \in \mathbb{Z}\}$$

3. The attempt at a solution

Don't I treat $$T_{1} \cup T_{2}$$ as a ring since any solution to the differential equation $$\left( \begin{array}{c}u^{'} & v^{'} \end{array} \right) = f(u,v) = 0$$ must lie on the Ring $$T_{1} \cup T_{2}$$?

Thus I must show $$T_{1} + T_{2}$$ and $$T_{1} \cdo T_{2}$$ lies
$$T_{1} \cup T_{2}$$?

If yes very well.

$$T_{1} + T_{2} = (\frac{4(l+1)\cdot \pi}{2}, \frac{4(m+1)\cdot \pi}{2})$$

and

$$T_{1} \cdot T_{2} = (\frac{(2 \cdot m+1)\cdot \pi}{2}, \frac{(2\cdot l+1)\cdot \pi}{2})$$

and since $$l,m \in \mathbb{Z}$$ then 2(l+1) and 2(m+1) and m(2m+1) and l(2l+1) are in $$T_{1} \cup T_{2}$$ is the Ring which defines in solution for $$\left( \begin{array}{c}u^{'} & v^{'} \end{array} \right) = f(u,v)$$ which implies that in l,m on the Ring mentioned above will give a solution for the differential equation.

How is that??

Sincerely Yours

Alphaboy

2. Oct 1, 2008

### tiny-tim

Hi Alphaboy! Thanks for the PM
I'm sorry, I'm not following this.

T1 and T2, according to your definitions, are the same.

And how can f(u,v) = 0? If the top line is 0, then the bottom line isn't.

And what is the differential equation?

And what is a direction field?

3. Oct 1, 2008

### Alphaboy2001

Hi Tiny-tim,

you are one of the smartest guys here :D

Anyway if look at my teachers definition of T1 and T2 its their inclusion which gives f(u,v) = 0 for any integer l and m, if and only if l and m are in Z.

and what I am trying to do it to insert T1 + T2 and T1*T2 into f(u,v) and as mentioned if l and m are in Z, then I get f(u,v) = 0.

and it was "vectorfield" and (u', v') is the corresponding differential equation.

I hope this have cleared it up now :)

Sincerely
Alpha